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Unformatted text preview: Math 218 (Spring 2008) Solutions to MT1 TA: Wei Lin 1. (a) Simply add up each row and column: 1 problem 2 problems 3 problems Total Brand A 20 12 3 35 Brand B 15 7 3 25 Brand C 10 6 4 20 Total 45 25 10 80 (b) P (Brand A and 1 problem) = 20 / 80 = 1 / 4. (c) The number of cars being Brand A or having 1 problem is the sum of all the numbers in the Brand A row and the 1 problem column, that is, 20 + 12 + 3 + 15 + 10 = 60. Hence, P (Brand A or 1 problem) = 60 / 80 = 3 / 4. Another method: P (Brand A or 1 problem) = P (Brand A) + P (1 problem) P (Brand A and 1 problem) = 35 / 80 + 45 / 80 20 / 80 = (35 + 45 20) / 80 = 3 / 4. (d) P (1 problem  Brand A) = # { 1 problem and Brand A } / # { Brand A } = 20 / 35 = 4 / 7. Another method: P (1 problem  Brand A) = P (1 problem and Brand A) /P (Brand A) = (20 / 80) / (35 / 80) = 20 / 35 = 4 / 7. (e) P (Brand A  1 problem) = # { Brand A and 1 problem } / # { 1 problem } = 20 / 45 = 4 / 9....
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This note was uploaded on 05/04/2008 for the course MATH 218 taught by Professor Haskell during the Fall '06 term at USC.
 Fall '06
 Haskell
 Math, Probability

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